)) CLASSROOM CONTACT PROGRAMME JEE(Main) UNIT …

)1001CJA102119029)(1001CJA102119029) Test Pattern

CLASSROOM CONTACT PROGRAMME(Academic Session : 2019 - 2020)

JEE(Main+Advanced) : ENTHUSIAST COURSE (SCORE-I)

Your Target is to secure Good Rank in JEE(Main) 2020

JEE(Main)UNIT TEST05-11-2019

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA-324005

+91-744-2757575 [email protected] www.allen.ac.in

En

glis

h

Name of the Candidate (in Capitals)

Form Number : in figures

: in words

Centre of Examination (in Capitals) :

Candidate’s Signature : Invigilator’s Signature :

Do not open this Test Booklet until you are askedto do so.Read carefully the Instructions on this Test Booklet.

Important Instructions :

1. Immediately fill in the form number on this page of the Test Booklet with Blue/Black Ball PointPen. Use of pencil is strictly prohibited.

2. The candidates should not write their Form Number anywhere else (except in the specified space)on the Test Booklet/Answer Sheet.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 75 questions. The maximum marks are 300.

5. There are three parts in the question paper 1,2,3 consisting of Physics, Chemistry andMathematics having 25 questions in each subject and each subject having Two sections.

(i) Section-I contains 20 multiple choice questions with only one correct option.

Marking scheme : +4 for correct answer, 0 if not attempted and –1 in all other cases.

(ii) Section-II contains 5 Numerical Value Type questions

Marking scheme : +4 for correct answer and 0 in all other cases.

6. Use Blue/Black Ball Point Pen only for writting particulars/marking responses on Side–1 andSide–2 of the Answer Sheet. Use of pencil is strictly prohibited.

7. No candidate is allowed to carry any textual material, printed or written, bits of papers, mobilephone any electronic device etc, except the Identity Card inside the examination hall/room.

8. Rough work is to be done on the space provided for this purpose in the Test Booklet only.

9. On completion of the test, the candidate must hand over the Answer Sheet to the invigilator onduty in the Room/Hall. However, the candidate are allowed to take away this Test Bookletwith them.

10. Do not fold or make any stray marks on the Answer Sheet.

Paper : Physics, Chemistry &Mathematics

PAPER-2

ALLEN

Page 2/16 SPACE FOR ROUGH WORK 05112019Enthusiast Course/Score-I/Paper-2 1001CJA102119029

PART 1 - PHYSICSSECTION–I : (Maximum Marks : 80)

� This section contains TWENTY questions.� Each question has FOUR options (A), (B), (C)

and (D). ONLY ONE of these four options iscorrect.

� For each question, darken the bubblecorresponding to the correct option in the ORS.

� For each question, marks will be awarded inone of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option isdarkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases

1. The orbital radius of a hydrogen atom is0.847 nm. The electron makes a transitionfrom this state to ground state. Themaximum possible number of wavelengthsthat can be emitted are :(A) 1 (B) 2 (C) 4 (D) 6

2. The minimum kinetic energy of a hydrogenatom required to produce a photon aftercollision with another hydrogen atom inground state at rest is :(A) 6.8 eV (B) 3.4 eV(C) 20.4 eV (D) 21 eV

3. Two cylinders A and B of equal volume arefilled with equal masses of N2 and O2respectively. Cylinder A is kept at 300 K whileB is kept at 600 K, then(A) Average kinetic energy of N2 (per mole)

in A is equal to that of O2 (per mole) in B.(B) Molecules in cylinder B move twice as

fast as those of A.(C) Pressure in flask A is less than that of B.(D) Average velocity of N2 is equal to rms

velocity of O2, in the given conditions.4. The V-T graph shows some process for an

ideal gas. The INCORRECT statement is:

EA

C

DB

F

T

V

(A) In process CD, pressure continuouslyincreases.

(B) In process EF, pressure continuouslydecreases.

(C) Pressure remains constant in all processAB, CD and EF.

(D) Internal energy of gas increases in allthree processes.

TOPIC : Modern Physics, KTG & Thermodynamics,Calorimetry, Heat Transfer, Thermal Expansion,

Elasticity, Semiconductor & Communication, Refrigeration, Heat Pump & Entropy)

ALLEN

05112019 SPACE FOR ROUGH WORK Page 3/16Enthusiast Course/Score-I/Paper-2 1001CJA102119029

5. At t = 0, light of having photon flux1012 photons/s-m2 of energy 6eV per photon startfalling on a plate with work function 2.5 eV. Ifarea of plate is 2 × 10–4 m2 and for every105 photons one photoelectron is emitted,charge on the plate at t = 25 sec is:(A) 8 × 10–15 C (B) 4 × 10–15 C(C)12 × 10–15 C (D) 16 × 10–15 C

6. When N0 molecules of a radioactive nuclideare taken at t = 0, the activity reduces to Ain time t0. When 3N0 molecules of the samenuclide are taken, activity will become 3Aein time t equal to :Where, l is the decay constant of the nuclide.

(A) t0 (B) 01t +l

(C) 01t -l

(D) None of these

7. 12 identical rods made of same material arearranged in the form of a cube. Thetemperature of 'P' and 'R' are maintained at90°C and 30°C respectively. The temperatureof 'V' in steady state is :

P

S R

W V

QUT

(A) 65°C (B) 60°C (C) 20°C (D) 50°C

8. A spring block system with mass of block mand spring constant k is placed on a smoothhorizontal plane as shown in the diagram. Thesurface area of the block is A. A light beam ofintensity I is switched on from rightwards.Assuming completely reflective surface, theamplitude of oscillations of the block is :(c speed of light in vaccum)

kI

(A) IAKC (B)

2IAKC (C)

4IAKC (D) Zero

9. Choose the CORRECT statement :(A)The message in an amplitude modulated

wave is contained in upper side bandonly.

(B)The message in an amplitude modulatedwave is contained in upper side band aswell as lower side band.

(C)The message in an amplitude modulatedwave is contained in carrier wave only.

(D)The message in an amplitude modulatedwave is contained in lower side bandonly.

ALLEN


10. The half life of a radioactive substance A is2 hour and that of B is 4 hour. The ratio ofactivity after 12 hours to initial activity of anequimolar mixture of A and B is :

(A) 9

64 (B) 5

96

(C) 1

16 (D) 1

16 2

11. A stone of mass m is tied to one end of a threadof length l. The diameter of thread is 'd' and itis suspended vertically. The stone is nowrotated in horizontal plane and makes anangle q with the vertical. The increase inlength of thread is (Young's modulus of threadis Y)

(A) 24mg

Y d cosp ql

(B) 24mg

Y d sinp ql

(C) 24mg

d Ypl

(D) 24mg

d Ysecp ql

12. An atom of atomic number z = 11 emits ka

wavelength which is l. The atomic number foran atom that emits ka radiation withwavelength 4l, is :(A) 2 (B) 4 (C) 6 (D) 8

13. One mole of an ideal gas is enclosed in acylinder fitted with a frictionless piston andoccupies a volume of 1.5 l at pressure of1.2 atm. It is subjected to a process given byequation T = aV2, the adiabatic constant forthe gas is g = 1.5. The INCORRECT statementis :

(A) The P-V graph is a straight line.

(B) The work done by gas in increasing thevolume of gas to 9 litre is 3150 J.

(C) The change in internal energy of the gasin increasing the volume of gas to 9 litre is12600 J

(D)The heat supplied to the gas in increasingthe volume of gas to 9 litre is 15730 J.

(Given that Ra = 80 J/mol-lit2, where R is thegas constant and a is constant)

14. The reading of a barometer containing someair above mercury column is 73 cm while thatof a correct one is 76 cm. If the tube of thefaulty barometer is pushed down into mercuryinitial volume of air in it is reduced to half,the reading shown by it will be :(A) 70 cm (B) 72 cm(C) 74 cm (D) 76 cm

ALLEN


15. The electric field of a light wave at a pointis E = (100 N/C) sin (3 × 1015 t) sin (6 × 1015t)where t is in seconds. The light falls on a metalsurface having work function (2eV), thenmaximum possible kinetic energy ofphotoelectrons is about :(A) 16 eV (B) 7 eV(C) 6 eV (D) 4 eV

16. Soon after Earth was formed, heat releasedby the decay of radioactive elements raised theaverage internal temperature from 300 to3000 K, at about which value it remains today.Assuming an average coefficient of volumeexpansion of 3.0 × 10–5 K–1, by how much hasthe radius of Earth increased since the planetwas formed?(A) 1.7 × 102

km (B) 2.4 × 102 km

(C) 1.7 × 103 km (D) 2.4 × 101

km17. The speeds of 22 particles are as follows

(Ni represents the number of particles thathave speed vi) :

( )i

i

N 2 4 6 8 2v cm / s 1.0 2.0 3.0 4.0 5.0

Which of the following statement isCORRECT?

(a) The average speed of all particles is

3.2 cm/sec.

(b) The rms speed of all particles is

3.4 cm/sec.

(c) The most probable speed of all particles is

4 cm/sec.

(d) The most probable speed of all particles is

5 m/s.

(A) (a) (B) (a) and (b)

(C) (a), (b) and (c) (D) All are correct

18. Mark the CORRECT statement :

(A) The nuclear force between two nucleons

depends upon the charge on each nucleon.

(B) The nuclear force is not a central force.

(C) The nuclear forces between nucleons

increases as the separation between the

nucleons increases.

(D) The nuclear force is independent of the spin

of the nucleons.

ALLEN


19. The rubidium isotope 8737 Rb , a b emitter with

a half life of 4.75 × 1010 yr, is used to determine

the age of rocks and fossils. Rocks containing

fossils of ancient animals contain a ratio of8738Sr to 87

37 Rb of 0.0160. Assuming that there

was no 8738Sr present when the rocks were

formed, estimate the age of these fossils.

(A) 2 × 108 year (B) 1 × 1012 year

(C) 1.1 × 109 year (D) 5 × 1011 year

20. A "Carnot" refrigerator (the reverse of a

Carnot engine) absorbs heat from the

freezer compartment at a temperature of

–17°C and exhausts it into the room at 25°C.

How much work must be done by the

refrigerator to change 0.50 kg of water at

27°C into ice at –23°C ?

(Take sW = 4200 J/kg–°C, Lf = 3.33 × 105 J/kg,

Sice = 2100 J/kg–°C).

(A) 3.9 × 104 J

(B) 4.9 × 104 J

(C) 1.2 × 105 J

(D) 8.4 × 105 J

SECTION-II : (Maximum Marks: 20)� This section contains FIVE questions.� The answer to each question is a

NUMERICAL VALUE.� For each question, enter the correct numerical

value (If the numerical value has more thantwo decimal places, truncate/round-off thevalue to TWO decimal places; e.g. 6.25, 7.00,–0.33, –.30, 30.27, –127.30, if answer is11.36777..... then both 11.36 and 11.37 willbe correct) by darken the correspondingbubbles in the ORS.For Example : If answer is –77.25, 5.2 thenfill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

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•

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•

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•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� Answer to each question will be evaluatedaccording to the following marking scheme:

Full Marks : +4 If ONLY the correctnumerical value is entered as answer.

Zero Marks : 0 In all other cases.

ALLEN


1. Two blocks of mass m and M are connectedby means of a metal wire passing over africtionless f ixed pulley. The area ofcross-section of wire is 6.5 × 10–9 m2 and itsbreaking stress is 2 × 109 N/m2. If m = 1kg,the maximum value of M (in kg) for whichthe wire will not break, is (g = 10 m/s2)

2. As a runner's foot pushes off the ground, theshear force acting on a 8 mm thick sole, isshown in the diagram. If the force of 25N isdistributed over the area of 15 cm2, the angleof shear is : (Given that shear modulus of thesole is 1.9 × 104 N/m2)

25N25N

q

3. In the circuit shown in figure, Zener diodeis properly biased. Power dissipated indiode (in mW) is :

V = 4VX

I2

I1

400W

400W12V

4. The radio nuclide decays according to11C ® 11B + e+ + u

The disintegration energy Q of this process(in MeV) is.Given that atomic masses mc=11.011433 u,me

= 0.0005486 u, mB = 11.009305 u,1 amu = 931 MeV

5. The change in entropy when a 30 g ice cube at–12°C is transformed into water at

100°C (in J/K) is (Take 273n 0.04261

æ ö =ç ÷è ø

l ,

373n 0.312273

æ ö =ç ÷è ø

l , Sice = 2100 J/kg-°C,

Swater = 4200 J/kg-°C, Lf = 3.33 × 105 J/kg)

ALLEN


SECTION–I : (Maximum Marks : 80)� This section contains TWENTY questions.� Each question has FOUR options (A), (B),

(C) and (D). ONLY ONE of these fouroptions is correct.

� For each question, darken the bubblecorresponding to the correct option in theORS.

� For each question, marks will be awardedin one of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option isdarkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases

1. Colloids can be purifed by which offollowing method-(A) Condensation (B) Peptisation(C) Coagulation (D) Dialysis

2. Which of the following statement is correct -(A) On adding excess of AgNO3 solution to

KI solution, negative colloid is formed(B) Ultra centrifugation process is used for

preparing hydrophobic colloids(C) Milk can be coagulated by adding acidic

or basic solution to it.(D) Protein sols are not always positive sol

3. Identify incorrect statement -(A) A catalyst cannot effect the position of

equilibrium of a reaction(B) A catalyst may change pre-exponential

factor of a reaction(C) Heterogeneous catalytic reactions are

always zero ordered reactions(D) It is possible that a species may increases

activation energy of a reaction

4.

Cl

NO2

O N2 NO2

(1)KOH(2)H+¾¾¾¾® X

Correct option regarding X?(A) X is more acidic than H2CO3

(B) X has alcoholic OH group(C) IUPAC name of X is

1-hydroxy-2, 4,6-trinitrobenzene(D) Formation of X is example of

electrophilic aromatic substitution

5. RCH = O 3

3

1.NH2.HCN Base3.H O+

+¾¾¾¾¾¾¾® RCH – NH3

CO2–

+

the amino acid which can not be preparedin this way is(A) Glycine (B) Alanine(C) Valine (D) Leucine

PART 2 - CHEMISTRY

TOPIC : Inorganic- Chemical bonding ; Organic- Aromatic Compounds, Biomolecules/Amino Acids , Polymer,Practical Organic Chemistry chemistry in everyday life ; Physical- Liquid Solution, Surface Chemistry

ALLEN


6.C

SNH

OO

O

For this compound if Lassaigne test iscarried out, what is the observation?(A) Blue colour is observed(B) Purple colour precipitate is observed(C) White precipitate is found which is

soluble in NH3 (excess)(D) Blood red colour is observed.

7. Which of the following have highest numberof 90° bond angle?(A) SF6 (B) IF7

(C) PF5 (D) 4ICl-

8. Which of the following have intra molecularH-bonding and found as dimer form invapour phase.(A) H2O(B) Acetic acid(C) Fumaric acid(D) none of these

9. Which of the following oxy acid containmaximum number of ionizable H-atom?(A) Pyrosulphuric acid(B) Pyrophosphoric acid(C) Orthophosphoric acid(D) Pyrophosphorous acid

10. At 300K, vapour pressure of pure toluene &pure benzene are 40 and 120 torrrespectively. For a solution which boils at300K and 50 torr, % w/w of benzene insolution is-(A) 10.8% (B) 21.6%(C) 32.6% (D) 89.2%

11. For an aqueous solution of 0.1M Ba(NO3)2 ,osmotic pressure at 300K is found to be6.576 atm. Percentage dissociation ofBa(NO3)2 in solution is nearly-(A) 91.3% (B) 84%(C) 74% (D) 100%

12. Which of the following is not example ofartifical sweetner?(A) Alitame (B) Aspartame(C) Sucrolose (D) Chloroxylenol

13. Which of the following heterocyclic base isnot present in DNA?(A) Guanine (B) Adenine(C) Quinoline (D) Cytosine

14. Which of the following contain maximumionic character according to Hanny andSmith formula?(A) HF (B) HCl(C) HBr (D) HI

15. Which of the following does not form duringthe hydrolysis of XeF6?(A) XeOF4 (B) XeO3

(C) XeO2F2 (D) XeO4

ALLEN


16. A solution contains 31 gm of ethylene glycolin 200 gm of water. If this solution is cooledto –5.4ºC. what mass of ice will separate outat this temperature ?[Kf for water = 1.8 K-kgmol–1](A) 33.3 g (B) 37.1 g(C) 48.2 g (D) 62.1 g

17.

NO2

NO2

3 22

1. NH H S2. NaNO HCl (0 5 C)3.CuCl HCl

++ - °

+

¾¾¾¾¾¾¾¾¾® Final

ProductFinal Product is?

(A)

Cl

(B)

Cl

Cl

(C)

NO2

Cl

(D)

Cl

ClO N2

18. Which of the following species isparamagnetic and has bond order of 2.5

(A) 2NÅ

(B) O2

(C) 22O -

(D) 2CÅ

19.

NH2

22 4

32 4

–

1. Ac O/ Pyridine2. Conc. H SO3. HNO4. dil. H SO /5. OH

D

¾¾¾¾¾¾¾® Product

Molecular formula of the major product is?

(A) C6H5N2SO3

(B) C6H6N2O2

(C) C8H8N2O3

(D) C6H5NSO5

20. Which of the following allotrope of carbondoes not contain dangling bond?

(A) Graphite

(B) Diamond

(C) Fullerene

(D) None of these

ALLEN



NUMERICAL VALUE.� For each question, enter the correct

numerical value (If the numerical value hasmore than two decimal places, truncate/round-off the value to TWO decimal places;e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ifanswer is 11.36777..... then both 11.36 and11.37 will be correct) by darken thecorresponding bubbles in the ORS.For Example : If answer is –77.25, 5.2 thenfill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

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••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

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••

� Answer to each question will be evaluatedaccording to the following marking scheme:Full Marks : +4 If ONLY the correctnumerical value is entered as answer.Zero Marks : 0 In all other cases.

1. 0.5 M aq. solution of a weak base BOH isprepared. Find elevation of boiling point(DTb) of this solution ( in K) given.For BOH ; Kb = 2.5 × 10–2 [ionisationconstant]For Water ; Kb = 5.2 K – Kg mol–1

[Ebullioscopic constant]Assume [Molality ; Molarity]

2. Oxygen gas is maintained at partial pressureof 0.54 atm above water sample. Henry's con-stant for dissolution of O2 in water is 105 atm.If molality of O2 in water at equilibrium is(y × 10–x) final value of 'x'

3. Find the total number of allotropic form is/are molecular solid.

White phosphorous, Red phosphorous,Black phosphorous, Rhombic sulphur,Monoclinic sulphur, Diamond, Graphite,Fullerene.

4. How many aldose when reacts with excessAc2O / pyridine increases molecular weightby 168 unit?

(A) Xylose

(B) Galactose

(C) Glucose

(D) Erythrose

(E) Lyxose

(F) Ribose

(G) Arabinose

5. Find the total number of species havepermanant dipole moment.

CH3Cl, CH2Cl2, CHCl3, CCl4,PF2Cl3, PF3Cl2,PF4Cl, SO2, SO3

ALLEN


PART 3 - MATHEMATICSSECTION–I : (Maximum Marks : 80)

� This section contains TWENTY questions.� Each question has FOUR options (A), (B),




1. Let <an>, <bn>, <cn> where n Î N arearithmetic progressions, such that

a1 + b1 + c1 = 10 and a2 + b2 + c2 = 20 then thevalue of a2019 + b2019 + c2019 equals to

(A) 20180 (B) 20190(C) 2018 (D) 2019

2. Equation x2 – 2(k – a)x + k2 + a2 – 16k – b +12 = 0 has repeated roots " k Î R, then valueof (a + b) is :-(A) 8 (B) 12(C) 16 (D) 20

3. The product of roots of 2log2x 4x 4 0l- + = is8 then l is :-(A) 2 (B) 2 2

(C) 3 (D) 3

4. If ( )n

2n r

r 1

1S t n 2n 9n 136=

= = + +å then value

of =

ån

rr 1

t is

(A) ( )+n n 1

2 (B) ( )+n n 2

2

(C) ( )+n n 3

2 (D) ( )+n n 5

25. A geometric progression has four

consecutive positive terms a1, a2, a3, a4. If

=3

1

a 9a

and + =1 24a a3 , then a4 equals to

(A) 3 (B) 9 (C) 27 (D) 3 36. If 3p2 = 5p + 2 and 3q2 = 5q + 2, then the

equation whose roots are 3p – 2q and 3q – 2pis(A) x2 – 5x + 100 = 0(B) 3x2 – 5x – 100 = 0(C) 3x2 + 5x + 100 = 0(D) 5x2 – x + 7 = 0

TOPIC : Complex number, Matrices, Determinants, Quadratic Equations, Sequence & Series

ALLEN


7. If a, b are roots of equation x2 – 2x – 1 = 0,then value of 5a4 + 12b3 is(A) 153 (B) 169 (C) 183 (D) 168

8. If a, b, c are in H.P. then - -b b ba , , c2 2 2

are

in :-(A) A.P. (B) G.P. (C) A.G.P.(D) H.P.

9. If M and N are two invertible skewsymmetric matrices such that MN = NM then(NTM–1N–1)T is(A) M (B) M–1 (C) N (D) N–1

10. a and b are roots of equation x2 – 2x + 4 = 0where Im(a) < 0 then value of a15 + b10.z0 is

(where 2i3

0z e ,i 1p

= = - )(A) 210 – 215 (B) 215 – 210

(C) 0 (D) 111. If < li , mi, ni > are direction cosines of three

mutually perpendicular straight lines for

i = 1,2,3 and 1 1 1

2 2 2

3 3 3

m nL m n

m n

é ùê ú= ê úê úë û

l

l

l

, then absolute

value of det(2L–1) is

(A) 8 (B) 18 (C) 4 (D)

14

12. Let A and B are square matrices of order3 × 3 such that AB = I, adjA = B – A, then(A) |B| > 1 (B) 0 < |B| < 1(C) |A| > 1 (D) |A| < 0(where |X| denotes determinant value ofmatrix X)

13. Let A be a square matrix of order 2 × 2 suchthat tr(A) = –2, A2 = I, then |A–1 + A2| is(where |X| and tr(X) denote determinantvalue and trace of matrix X respectively)(A) 1(B) –1(C) 0(D) can't be determined

14. 2 21 1 1

1 1+ +

- w w - w w - is equal to (w is non

real cube root of unity)(A) 0 (B) 1 (C) –1 (D) w

15. If ( ) ( )n

2r 1

1 10r r 1 1 10=

w=

w - w - + - wå

then n is 2i3where e ,i 1pæ ö

ç ÷w = = -ç ÷è ø

(A) 9 (B) 10 (C) 8 (D) 1116. The system of equations

ax + y + z = ax + ay + z = ax + y + az = ahas no solution, if(A) a = –2, only (B) a = –2, – 1 only(C) a = –2,1 only (D) a Î R – {–2}

17. Let A = [aij]3×3 and C = [cij]3×3 " i, j Î {1,2,3}be two square matrices, where cij is cofactorof aij in |A|such that 3cij – 2aij = 0 andATA = lI, l ¹ 0, then value of l is

(A) 23 (B)

49 (C)

94 (D)

32

ALLEN


18. If

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( )

2 22 2 3

2 2 2 6 22 3 4 5

2 2 23 4 5

1 1 1

1 1 1 1 1

1 1 1

a + a + a +

a + a + a + = la - a + a

a + a + a +

,

then l is equal to

(A) 1 (B) 4 (C) 2 (D) –2

19. The following system of linear equations

x + y + z = 1

x + ay + z = 1

x + by + az = 0

has infinite solutions, then

(A) b has only one value

(B) b Î f

(C) b Î R

(D) b Î R – {1}

20. If a,b,g are the roots of the equation

x3+ax2 + b = 0, b ¹ 0 and

21 1 1

1 1 1

1 1 1

a b g

D =b g a

g a b

, then

D is equal to(A) a2 – b (B) b2 – a

(C) a2b2 (D) 0




0 0 0

+

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–

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+

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0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

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ALLEN


1. Let 1 0 0

A 2 1 02 2 1


, then det(A10 + A–10) is

equal to

2. Radius of 1Re 1i z

æ ö =ç ÷+è ø is r, then 3r

2 is equal to

(where z = x + iy, i 1= - , x,y Î R)

3. Let z1 and z2 are two complex numbers such

that 1 2z z 1= and arg(z1) – arg(z2) = 23p

, then

value of 4i3

1 2z z ep

+ is equal to

(where i 1= - , –p < arg(z) £ p)

4. All the points in the set

t 2iS ,t R,i 1t 2i

+ì ü= Î = -í ý-î þlie on the curve

ƒ(z) = 0, z = x + iy where x, y Î R thenmaximum value of |z + i – 2| is

5. If z1 and z2 are non zero solutions of equation2z z iz+ = where i 1= - , then

|z1 + z2| is equal to

ALLEN


SPACE FOR ROUGH WORK

Page 16/16 05112019Enthusiast Course/Score-I/Paper-2 1001CJA102119029

Name of the Candidate (in Capitals)ijh{kkFkhZ dk uke (cM+s v{kjksa esa) :Form Number : in figuresQkWeZ uEcj : vadksa esa

: in words: 'kCnksa esa

Centre of Examination (in Capitals) :ijh{kk dsUæ (cM+s v{kjksa esa) :Candidate’s Signature : Invigilator’s Signature :ijh{kkFkhZ ds gLrk{kj : fujh{kd ds gLrk{kj :

Read carefully the Instructions this Test Booklet.bl ijh{kk iq fLrdk ij fn, funsZ'kk s a dks /;ku ls i<+ s aA

Do not open this Test Booklet until you are asked to do so.bl ijh{kk iqfLrdk dks tc rd uk [kk sysa tc rd dgk u tk,A

)1001CJA102119029)(1001CJA102119029) Test Pattern

egRoiw.kZ funs Z'k :1. ijh{kk iqfLrdk ds bl i`"B ij vko';d fooj.k uhys@dkys ckWy ikbaV

isu ls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gSaA

2. ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk

iqfLrdk@mÙkj i= ij dgha vkSj u fy[ksaA

3. ijh{kk dh vof/k 3 ?kaVs gSA4. bl ijh{kk iqfLrdk esa 75 iz'u gaSA vf/kdre vad 300 gSaA

5. bl ijh{kk iq fLrdk esa rhu Hkkx 1, 2, 3 gSa] ftlds izR;sd Hkkx esa HkkSfrdfoKku] jlk;u foKku ,oa xf.kr ds 25 iz'u gSa vkSj izR;sd fo"k;esa 2 [k.M gSA(i) [k.M-I esa 20 cgqfodYih; iz'u gSA ftuds dsoy ,d fodYi

lgh gSaAvad ;k stuk : +4 lgh mÙkj ds fy,] 0 iz;kl ugha djus ijrFkk –1 vU; lHkh voLFkkvksa esaA

(ii) [k.M-II esa 5 la[;kRed eku izdkj ds iz'u gSAvad ;kstuk : +4 lgh mÙkj ds fy, rFkk 0 vU; lHkh voLFkkvksa esaA

6. mÙkj i= ds i`"B–1 ,oa i`"B–2 ij okafNr fooj.k ,oa mÙkj vafdr djusgsrq dsoy uhys@dkys ckWy ikbaV isu dk gh iz;ksx djsaA isfUly dk iz;ksxloZFkk oftZr gSA

7. ijh{kkFkhZ }kjk ijh{kk d{k @ gkWy esa ifjp; i= ds vykok fdlh Hkh izdkjdh ikB~; lkexzh eqfær ;k gLrfyf[kr dkxt dh ifpZ;ksa] eksckbyQksu ;k fdlh Hkh izdkj ds bysDVªkfud midj.kksa ;k fdlh vU; izdkjdh lkexzh dks ys tkus ;k mi;ksx djus dh vuqefr ugha gSaA

8. jQ dk;Z ijh{kk iqfLrdk esa dsoy fu/kkZfjr txg ij gh dhft;sA

9. ijh{kk lekIr gksus ij] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ mÙkj i= d{kfujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk bl ijh{kk iq fLrdkdks ys tk ldrs gSaA

10. mÙkj i= dks u eksM+ s a ,oa u gh ml ij vU; fu'kku yxk,s aA

Important Instructions :1. Immediately fill in the form number on this page of the

Test Booklet with Blue/Black Ball Point Pen. Use of pencilis strictly prohibited.

2. The candidates should not write their Form Numberanywhere else (except in the specified space) on the TestBooklet/Answer Sheet.

3. The test is of 3 hours duration.4. The Test Booklet consists of 75 questions. The maximum

marks are 300.5. There are three parts in the question paper 1,2,3

consisting of Physics, Chemistry and Mathematicshaving 25 questions in each subject and each subjecthaving Two sections.(i) Section-I contains 20 multiple choice questions

with only one correct option.Marking scheme : +4 for correct answer, 0 if notattempted and –1 in all other cases.

(ii) Section-II contains 5 Numerical Value TypequestionsMarking scheme : +4 for correct answer and 0 inall other cases.

6. Use Blue/Black Ball Point Pen only for writtingparticulars/marking responses on Side–1 and Side–2 of theAnswer Sheet. Use of pencil is strictly prohibited.

7. No candidate is allowed to carry any textual material,printed or written, bits of papers, mobile phone anyelectronic device etc, except the Identity Card inside theexamination hall/room.

8. Rough work is to be done on the space provided for thispurpose in the Test Booklet only.

9. On completion of the test, the candidate must hand overthe Answer Sheet to the invigilator on duty in the Room/Hall. However, the candidate are allowed to take awaythis Test Booklet with them.

10. Do not fold or make any stray marks on the Answer Sheet.

CLASSROOM CONTACT PROGRAMME(Academic Session : 2019 - 2020)

JEE(Main+Advanced) : ENTHUSIAST COURSE (SCORE-I)

Your Target is to secure Good Rank in JEE(Main) 2020

JEE(Main)UNIT TEST05-11-2019

Paper : Physics, Chemistry & Mathematicsç'u iq fLrdk : HkkSfrd foKku] jlk;u foKku rFkk xf.kr

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA-324005

+91-744-2757575 [email protected] www.allen.ac.in

PAPER-2

Hin

di

ALLEN

Page 2/28 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 05112019Enthusiast Course/Score-I/Paper-2 1001CJA102119029

SECTION–I : (Maximum Marks : 80)� This section contains TWENTY questions.� Each question has FOUR options (A), (B), (C)

and (D). ONLY ONE of these four options iscorrect.

� For each question, darken the bubblecorresponding to the correct option in the ORS.

� For each question, marks will be awarded inone of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option isdarkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases

1. The orbital radius of a hydrogen atom is0.847nm. The electron makes a transitionfrom this state to ground state. Themaximum possible number of wavelengthsthat can be emitted are :(A) 1 (B) 2 (C) 4 (D) 6

2. The minimum kinetic energy of a hydrogenatom required to produce a photon aftercollision with another hydrogen atom inground state at rest is :(A) 6.8 eV (B) 3.4 eV(C) 20.4 eV (D) 21 eV

[k.M–I : (vf/kdre vad : 80)

� bl [k.M esa chl iz'u gSa

� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D)

gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi ds

vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls

fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +4 ;fn flQZ lgh fodYi ds vuq:i cqycqys

dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

1. ,d gkbMªkstu ijek.kq dh d{kh; f=T;k 0.847 nm gSA

bysDVªkWu bl voLFkk ls ewy voLFkk eas laØe.k djrk gSA

mRlftZr gks ldus okyh rjaxnS/;ks± dh vf/kdre laHkkfor

la[;k gksxh :(A) 1 (B) 2 (C) 4 (D) 6

2. ,d gkbMªkstu ijek.kq dh ewy voLFkk esa fojkekoLFkk eas

fLFkr vU; gkbMªkstu ijek.kq ls VDdj ds i'pkr~ QksVksu

mRiUu djus ds fy, vko';d U;wure xfrt ÅtkZ gksxh :(A) 6.8 eV (B) 3.4 eV(C) 20.4 eV (D) 21 eV

PART 1 - PHYSICS

TOPIC : Modern Physics, KTG & Thermodynamics,Calorimetry, Heat Transfer, Thermal Expansion,

Elasticity, Semiconductor & Communication, Refrigeration, Heat Pump & Entropy)

ALLEN

05112019 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 3/28Enthusiast Course/Score-I/Paper-2 1001CJA102119029

3. Two cylinders A and B of equal volume arefilled with equal masses of N2 and O2

respectively. Cylinder A is kept at 300 Kwhile B is kept at 600 K, then(A) Average kinetic energy of N2 (per mole)

in A is equal to that of O2 (per mole) in B.(B) Molecules in cylinder B move twice as

fast as those of A.(C) Pressure in flask A is less than that of B.(D) Average velocity of N2 is equal to rms

velocity of O2, in the given conditions.4. The V-T graph shows some process for an

ideal gas. The INCORRECT statement is:

EA

C

DB

F

T

V

(A) In process CD, pressure continuouslyincreases.

(B) In process EF, pressure continuouslydecreases.

(C) Pressure remains constant in all processAB, CD and EF.

(D) Internal energy of gas increases in allthree processes.

3. leku vk;ru okys nks csyuksa A rFkk B dks Øe'k% N2

rFkk O2 ds leku æO;eku ls Hkjk tkrk gSA csyu A dks

300 K ij j[kk tkrk gS] tcfd csyu B dks 600 K ij

j[kk tkrk gS] rks

(A) A esa N2 (çfr eksy) dh vkSlr xfrt ÅtkZ B esa

O2 (çfr eksy) dh vkSlr xfrt ÅtkZ ds leku gksxhA

(B) csyu B esa v.kq csyu A ds v.kqvksa dh rqyuk esa

nqxquh rsth ls xfr djrs gSA

(C) ¶ykLd A esa nkc ¶ykLd B dh rqyuk esa de gSA

(D) nh xbZ fLFkfr;ksa esa N2 dk vkSlr osx O2 ds oxZ ekè;

ewy osx ds leku gSA

4. iznf'kZr V-T vkjs[k esa ,d vkn'kZ xSl ds fy, dqN çØe

n'kkZ;s x;s gSA xyr dFku pqfu;sA

EA

C

DB

F

T

V

(A) çØe CD esa nkc yxkrkj c<+rk gSA

(B) çØe EF esa nkc yxkrkj ?kVrk gSA

(C) lHkh çØekas AB, CD rFkk EF esa nkc fu;r cuk

jgrk gSA

(D) lHkh rhuksa çØeksa esa xSl dh vkUrfjd ÅtkZ c<+rh gSA

ALLEN


5. le; t = 0 ij dk;ZQyu 2.5 eV okyh ,d IysV ijçfr QksVksu 6eV ÅtkZ dk QksVkWu ¶yDl 1012 QksVkWu/s-m2

okyk çdk'k vkifrr gksuk çkjEHk gksrk gSA ;fn IysV dk{ks=Qy 2 × 10–4 m2 gks rFkk çR;sd 105 QksVksu ds fy,,d çdk'kbysDVªkWu mRlftZr gksrk gS rks t = 25 sec ijIysV ij vkos'k gksxk :(A) 8 × 10–15 C (B) 4 × 10–15 C(C)12 × 10–15 C (D) 16 × 10–15 C

6. tc t = 0 ij jsfM;kslfØ; U;wDykbM ds N0 v.kq fy;stkrs gS rks t0 le; esa lfØ;rk A rd ?kV tkrh gSA ;fnblh U;wDykbM ds 3N0 v.kq fy;s tkrs gS rks lfØ;rkfdrus le; t esa 3Ae gks tk;sxh\tgk¡ l U;wDykbM dk {k; fu;rkad gSA

(A) t0 (B) 01t +l

(C) 01t -l

(D) buesa ls dksbZ ughaA

7. leku inkFkZ ls cuh 12 ,dtSlh NM+ksa dks ?ku ds :i esaO;ofLFkr fd;k tkrk gSA 'P' rFkk 'R' dk rkieku Øe'k%90°C rFkk 30°C j[kk tkrk gSA LFkk;h voLFkk esa 'V' dkrkieku gS :

P

S R

W V

QUT

(A) 65°C (B) 60°C (C) 20°C (D) 50°C

5. At t = 0, light of having photon flux1012 photons/s-m2 of energy 6eV per photonstart falling on a plate with work function2.5 eV. If area of plate is 2 × 10–4 m2 and forevery 105 photons one photoelectron isemitted, charge on the plate at t = 25 sec is:(A) 8 × 10–15 C (B) 4 × 10–15 C(C)12 × 10–15 C (D) 16 × 10–15 C

6. When N0 molecules of a radioactive nuclideare taken at t = 0, the activity reduces to A intime t0. When 3N0 molecules of the samenuclide are taken, activity will become 3Ae intime t equal to :Where, l is the decay constant of the nuclide.

(A) t0 (B) 01t +l

(C) 01t -l

(D) None of these

7. 12 identical rods made of same material arearranged in the form of a cube. Thetemperature of 'P' and 'R' are maintained at90°C and 30°C respectively. The temperatureof 'V' in steady state is :

P

S R

W V

QUT

(A) 65°C (B) 60°C (C) 20°C (D) 50°C

ALLEN


8. æO;eku m okys CYkkWd rFkk fLçax fu;rkad k okyh fLçax

ds (fLçax CykWd) fudk; dks ,d fpdus {kSfrt ry ij

fp=kuqlkj j[kk tkrk gSA CykWd dk i`"Bh; {ks=Qy A gSA

rhozrk I okys ,d çdk'k iq¡t dks nk¡;h vksj ls vkWu fd;k

tkrk gSA iw.kZ:i ls ijkorZd lrg ekurs gq;s CykWd ds

nksyuksa dk vk;ke gS : (c fuokZr esa çdk'k dh pky gSA)

kI

(A) IAKC (B)

2IAKC (C)

4IAKC (D) 'kwU;

9. lgh dFku pqfu;s%&

(A) fdlh vk;ke ekWMwfyr rjax esa lwpuk dsoy Åijh

ik'oZ cS.M esa gh fo|eku gksrh gSA

(B) fdlh vk;ke ekWMwfyr rjax esa lwpuk Åijh rFkk fupys

nksuksa ik'oZ cS.M esa fo|eku gksrh gSA

(C) fdlh vk;ke ekWMwfyr rjax esa lwpuk dsoy okgd

rjax esa gh fo|eku gksrh gSA

(D)fdlh vk;ke ekWMwfyr rjax esa lwpuk dsoy fupys

ik'oZ cS.M esa gh fo|eku gksrh gSA

8. A spring block system with mass of block mand spring constant k is placed on a smoothhorizontal plane as shown in the diagram.The surface area of the block is A. A lightbeam of intensity I is switched on fromrightwards. Assuming completely reflectivesurface, the amplitude of oscillations of theblock is : (c speed of light in vaccum)

kI

(A) IAKC (B)

2IAKC (C)

4IAKC (D) Zero

9. Choose the CORRECT statement :(A)The message in an amplitude modulated

wave is contained in upper side bandonly.

(B)The message in an amplitude modulatedwave is contained in upper side band aswell as lower side band.

(C)The message in an amplitude modulatedwave is contained in carrier wave only.

(D)The message in an amplitude modulatedwave is contained in lower side bandonly.

ALLEN


10. ,d jsfM;kslfØ; inkFkZ A dh v¼Zvk;q 2 ?kaVs rFkk B

dh v¼Zvk;q 4 ?kaVs gSA A rFkk B ds lerqY; feJ.k dh

12 ?kaVs i'pkr~ lfØ;rk rFkk izkjfEHkd lfØ;rk dk vuqikr

gksxk%&

(A) 9

64 (B) 5

96

(C) 1

16 (D) 1

16 2

11. æO;eku m okys ,d iRFkj dks yEckbZ l okys /kkxs ds ,d

fljs ls ck¡/kk x;k gSA /kkxs dk O;kl 'd' gS rFkk ;g Å/okZèkj

:i ls yVdk;k x;k gSA vc iRFkj dks {kSfrt ry esa

?kw.kZu djk;k tkrk gS rFkk ;g ÅèokZ/kj ls q dks.k cukrk

gSA /kkxs dh yEckbZ esa o`f¼ gS : (/kkxs dk ; ax çR;kLFkrk

xq.kkad Y gSA)

(A) 24mg

Y d cosp ql

(B) 24mg

Y d sinp ql

(C) 24mg

d Ypl

(D) 24mg

d Ysecp ql

12. ijek.kq Øekad z = 11 okyk ,d ijek.kq ka rjaxnS/;Z

mRlftZr djrk gS ftldk eku l gSA rjaxnS/;Z 4l okys

ka fofdj.k mRlftZr djus okys ijek.kq dk ijek.kq Øekad

gksxk :(A) 2 (B) 4 (C) 6 (D) 8

10. The half life of a radioactive substance A is2 hour and that of B is 4 hour. The ratio ofactivity after 12 hours to initial activity ofan equimolar mixture of A and B is :

(A) 9

64 (B) 5

96

(C) 1

16 (D) 1

16 2

11. A stone of mass m is tied to one end of athread of length l. The diameter of threadis 'd' and it is suspended vertically. The stoneis now rotated in horizontal plane and makesan angle q with the vertical. The increase inlength of thread is (Young's modulus ofthread is Y)

(A) 24mg

Y d cosp ql

(B) 24mg

Y d sinp ql

(C) 24mg

d Ypl

(D) 24mg

d Ysecp ql

12. An atom of atomic number z = 11 emits ka

wavelength which is l. The atomic numberfor an atom that emits ka radiation withwavelength 4l, is :(A) 2 (B) 4 (C) 6 (D) 8

ALLEN


13. ,d eksy vkn'kZ xSl ?k"kZ.kjfgr fiLVu ; qDr csyu esa

ifjc¼ gS rFkk 1.2 ok;qe.Myh; nkc ij 1.5 l vk;ru

?ksjrh gSA ;g xSl lehdj.k T = aV2 }kjk iznf'kZr izØe

dk vuqlj.k djrh gS rFkk xSl dk :¼ks"e ?kkrkad g = 1.5

gSA xyr dFku pqfu;sA

(A) P-V vkjs[k ljy js[kh; gSA

(B) xSl ds vk;ru dks 9 yhVj rd c<+kus esa xSl }kjk

fd;k x;k dk;Z 3150 J gSA

(C) xSl ds vk;ru dks 9 yhVj rd c<+kus esa xSl dh

vkUrfjd ÅtkZ esa ifjorZu 12600 J gSA

(D) xSl ds vk;ru dks 9 yhVj rd c<+kus esa xSl dks nh xbZ

Å"ek 15730 J gSA

(fn;k gS Ra = 80 J/mol-lit2, tgk¡ R xSl fu;rkad rFkk

a fu;rkad gSA)

14. ,d cSjksehVj esa ikjs ds LrEHk ds Åij dqN ok;q Hkjh gqbZ gS

rFkk ikB~;kad 73 cm gS tcfd lgh ikB~;kad 76 cm gSA

;fn = qfViw.kZ cSjksehVj dh uyh dks ikjs ds vUnj uhps dh

vksj /kdsyk tkrk gS] rks blesa ok;q dk çkjfEHkd vk;ru

vk/kk de gks tkrk gSA blds }kjk çnf'kZr ikB~;kad gksxk :

(A) 70 cm (B) 72 cm(C) 74 cm (D) 76 cm

13. One mole of an ideal gas is enclosed in acylinder fitted with a frictionless piston andoccupies a volume of 1.5 l at pressure of1.2atm. It is subjected to a process given byequation T = aV2, the adiabatic constant forthe gas is g = 1.5. The INCORRECTstatement is :

(A)The P-V graph is a straight line.

(B)The work done by gas in increasing thevolume of gas to 9 litre is 3150 J.

(C)The change in internal energy of the gasin increasing the volume of gas to 9 litreis 12600 J

(D)The heat supplied to the gas in increasingthe volume of gas to 9 litre is 15730 J.

(Given that Ra = 80 J/mol-lit2, where R isthe gas constant and a is constant)

14. The reading of a barometer containing someair above mercury column is 73 cm while thatof a correct one is 76 cm. If the tube of thefaulty barometer is pushed down intomercury initial volume of air in it is reducedto half, the reading shown by it will be :(A) 70 cm (B) 72 cm(C) 74 cm (D) 76 cm

ALLEN


15. fdlh fcUn q ij çdk'k rjax dk fo|qr {k s=

E = (100 N/C) sin (3 × 1015 t) sin (6 × 1015t) gS]

tgk¡ t lsd.M esa gSA çdk'k ] dk;ZQyu (2eV) okyh

èkkfRod lrg ij vkifrr gksrk gS rks çdk'kbysDVªksuksa dh

vf/kdre lEHkkfor xfrt ÅtkZ yxHkx gksxh :

(A) 16 eV

(B) 7 eV

(C) 6 eV

(D) 4 eV

16. i`Foh cuus ds rqjUr i'pkr~ jsfM;kslfØ; rRoksa ds fo?kVu

}kjk mRlftZr Å"ek ds dkj.k vkSlr vkUrfjd rkieku

300 ls 3000 K rd c<+k Fkk tks yxHkx vkt Hkh cuk

gqvk gSA vkSlr vk;ru çlkj xq.kkad 3.0 × 10–5 K–1

ekurs gq;s xzg ds fufeZr gksus ls vc rd i`Foh dh f=T;k

esa fdruh o`f¼ gq;h gS\

(A) 1.7 × 102 km

(B) 2.4 × 102 km

(C) 1.7 × 103 km

(D) 2.4 × 101 km

15. The electric field of a light wave at a pointis E = (100 N/C) sin (3 × 1015 t) sin (6 × 1015t)where t is in seconds. The light falls on ametal surface having work function (2eV),then maximum possible kinetic energy ofphotoelectrons is about :

(A) 16 eV

(B) 7 eV

(C) 6 eV

(D) 4 eV

16. Soon after Earth was formed, heat releasedby the decay of radioactive elements raisedthe average internal temperature from 300to 3000 K, at about which value it remainstoday. Assuming an average coefficient ofvolume expansion of 3.0 × 10–5 K–1, by howmuch has the radius of Earth increased sincethe planet was formed?

(A) 1.7 × 102 km

(B) 2.4 × 102 km

(C) 1.7 × 103 km

(D) 2.4 × 101 km

ALLEN


17. fp=kuqlkj 22 d.kksa dh pky nh xbZ gSA(Ni d.kksa dh

la[;k çnf'kZr djrk gS ftudh pky vi gS) :

( )i

i

N 2 4 6 8 2v cm / s 1.0 2.0 3.0 4.0 5.0

fuEu esa ls dkSulk dFku lgh gS\

(a) lHkh d.kksa dh vkSlr pky 3.2 cm/sec gSA

(b) lHkh d.kksa dh oxZ ek/; ewy pky 3.4 cm/sec gSA

(c) lHkh d.kksa dh vf/kdre lEHkkO; pky 4 cm/sec gSA

(d) lHkh d.kksa dh vf/kdre lEHkkO; pky 5 m/s gSA

(A) (a) (B) (a) rFkk (b)

(C) (a), (b) rFkk (c) (D) lHkh lgh gSA

18. lgh dFku pqfu;s :

(A) nks U;wfDy;kWuksa ds e/; yxus okyk ukfHkdh; cy

çR;sd U;wfDy;kWu ij fLFkr vkos'k ij fuHkZj djrk

gSA

(B) ukfHkdh; cy dsUæh; cy ugha gksrk gSA

(C) U;wfDy;kWuksa ds e/; yxus okys ukfHkdh; cy dk

eku U;wfDy;kWuksa ds e/; nwjh c<+us ij c<+rk gSA

(D) ukfHkdh; cy U;wfDy;kWuksa ds pØ.k ij fuHkZj ughdjrk gSA

17. The speeds of 22 particles are as follows (Ni

represents the number of particles that havespeed vi) :

( )i

i

N 2 4 6 8 2v cm / s 1.0 2.0 3.0 4.0 5.0

Which of the following statement isCORRECT?(a) The average speed of all particles is

3.2 cm/sec.(b) The rms speed of all particles is

3.4 cm/sec.(c) The most probable speed of all particles

is 4 cm/sec.(d) The most probable speed of all particles

is 5 m/s.(A) (a) (B) (a) and (b)(C) (a), (b) and (c) (D) All are correct

18. Mark the CORRECT statement :(A) The nuclear force between two nucleons

depends upon the charge on eachnucleon.

(B) The nuclear force is not a central force.(C) The nuclear forces between nucleons

increases as the separation between thenucleons increases.

(D) The nuclear force is independent of thespin of the nucleons.

ALLEN


19. :fcfM;e leLFk k fud 8737 Rb rFk k v¼Zvk; q

4.75 × 1010 o"kZ okys b mRltZd dks pV~Vkuksa rFkk

thok'eks a dh vk;q Kkr djus ds fy, ç;qDr fd;k tkrk

gSA çkphu tkuojksa ds thok'e ; qDr pV~Vkuksa esa 8738Sr

rFkk 8737 Rb dk vuqikr 0.0160 gSA ekuk fd tc pV~Vkusa

cuh Fkh rc dksbZ 8738Sr mifLFkr ugha Fkk] rks bu thok'eks a

dh vk;q Kkr dhft;sA

(A) 2 × 108 o"kZ

(B) 1 × 1012 o"kZ

(C) 1.1 × 109 o"kZ

(D) 5 × 1011 o"kZ

20. ,d dkuksZ jsfÝtjsVj (dkuksZ batu dk O;qRØe) –17°C

rkieku ij fÝtj ckWDl ls Å"ek vo'kksf"kr djrk gS rFkk

bls 25°C okys dejs esa mRlftZr djrk gSA jsfÝtjsVj

}kjk 27°C okys 0.50 kg ty dks –23°C okys cQZ esa

ifjofrZr djus ds fy;s fdruk dk;Z fd;k tkuk pkfg;s\

(sW = 4200 J/kg–°C, Lf = 3.33 × 105 J/kg,

Sice = 2100 J/kg–°C ysa)

(A) 3.9 × 104 J(B) 4.9 × 104 J(C) 1.2 × 105 J(D) 8.4 × 105 J

19. The rubidium isotope 8737 Rb , a b emitter with

a half life of 4.75 × 1010 yr, is used todetermine the age of rocks and fossils. Rockscontaining fossils of ancient animals contain

a ratio of 8738Sr to 87

37 Rb of 0.0160. Assuming

that there was no 8738Sr present when the

rocks were formed, estimate the age of thesefossils.

(A) 2 × 108 year

(B) 1 × 1012 year

(C) 1.1 × 109 year

(D) 5 × 1011 year

20. A "Carnot" refrigerator (the reverse of aCarnot engine) absorbs heat from thefreezer compartment at a temperature of–17°C and exhausts it into the room at 25°C.How much work must be done by therefrigerator to change 0.50 kg of water at27°C into ice at –23°C ?

(Take sW = 4200 J/kg–°C, Lf = 3.33 × 105 J/kg,Sice = 2100 J/kg–°C).

(A) 3.9 × 104 J

(B) 4.9 × 104 J

(C) 1.2 × 105 J

(D) 8.4 × 105 J

ALLEN


[kaM-II : (vf/kdre vad : 20)� bl [kaM esa ik¡p iz'u gSaA� izR;sd iz'u dk mÙkj ,d la[;kRed eku (NUMERICAL

VALUE) gSA� iz R; sd i z'u ds mÙkj ds lgh la[; kRed eku

(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuks a rdVª ad sV@jkm aM vk WQ (truncate/round-off) djs a_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaA

mnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk %&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA




0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� Answer to each question will be evaluatedaccording to the following marking scheme:

Full Marks : +4 If ONLY the correctnumerical value is entered as answer.

Zero Marks : 0 In all other cases.

ALLEN


1. Two blocks of mass m and M are connectedby means of a metal wire passing over a

frictionless f ixed pulley. The area of

cross-section of wire is 6.5 × 10–9 m2 and its

breaking stress is 2 × 109 N/m2. If m = 1kg,the maximum value of M (in kg) for which

the wire will not break, is (g = 10 m/s2)

2. As a runner's foot pushes off the ground, theshear force acting on a 8 mm thick sole, isshown in the diagram. If the force of 25N isdistributed over the area of 15 cm2, the angleof shear is : (Given that shear modulus ofthe sole is 1.9 × 104 N/m2)

25N25N

q

1. æO;eku m rFkk M okys nks CykWd ?k"kZ.kjfgr fLFkj f?kjuh

ij ls xqtjus okys ,d /kkfRod rkj ls tqM+s gq;s gSA rkj ds

vuqçLFk dkV dk {ks=Qy 6.5 × 10–9 m2 gS rFkk bldk

Hkatu çfrcy 2 × 109 N/m2 gSA ;fn m = 1kg gS rks

M (kg esa) dk og vf/kdre eku Kkr dhft;s ftlds

fy, rkj VwVsxk ughaA (g = 10 m/s2)

2. tc ,d /kkod vius iSj ls /kjkry dks nckrk gS rks

8mm eksVs twrs ds rys ij dk;Zjr vi:i.k cy fp=kuqlkj

fn;k tkrk gSA ;fn 25N cy 15 cm2 {kS=Qy ij forfjr

gksrk gS rks vi:i.k dks.k dk eku Kkr dhft;sA (rys dk

vi:i.k xq.kkad 1.9 × 104 N/m2 fn;k x;k gSA)

25N25N

q

ALLEN


3. In the circuit shown in figure, Zener diodeis properly biased. Power dissipated indiode (in mW) is :

V = 4VX

I2

I1

400W

400W12V

4. The radio nuclide decays according to11C ® 11B + e+ + u

The disintegration energy Q of this process(in MeV) is.Given that atomic masses mc=11.011433 u,me

= 0.0005486 u, mB = 11.009305 u,1 amu = 931 MeV

5. The change in entropy when a 30 g ice cubeat –12°C is transformed into water at

100°C (in J/K) is (Take 273n 0.04261

æ ö =ç ÷è ø

l ,

373n 0.312273

æ ö =ç ÷è ø

l , Sice = 2100 J/kg-°C,

Swater = 4200 J/kg-°C, Lf = 3.33 × 105 J/kg)

3. çnf'kZr ifjiFk esa tsuj Mk;ksM mfpr :i ls ck;flr

gSA Mk;ksM esa O;f;r 'kfDr dk eku (mW esa) Kkr

dhft;sA

V = 4VX

I2

I1

400W

400W12V

4. jsfM;ks U;wDykbM11C ® 11B + e+ + u

ds vuqlkj fo?kfVr gksrk gSA bl çØe dh fo?kVu ÅtkZ

Q dk eku MeV esa Kkr dhft;sA

fn;k g S % ijek.k q Øekad mc=11.011433 u,me

= 0.0005486 u, mB = 11.009305 u,1 amu = 931 MeV

5. tc –12°C okyk 30g cQZ dk ?ku 100°C okys ty eas

ifjofrZr gksrk gS rks ,UVªksih esa ifjorZu J/K esa Kkr dhft;sA

(273n 0.04261

æ ö =ç ÷è ø

l ,373n 0.312273

æ ö =ç ÷è ø

l ,

Sice = 2100 J/kg-°C, Swater = 4200 J/kg-°C,

L f = 3.33 × 105 J/kg ysa)

ALLEN






1. Colloids can be purifed by which offollowing method-(A) Condensation (B) Peptisation(C) Coagulation (D) Dialysis

2. Which of the following statement is correct -(A) On adding excess of AgNO3 solution to

KI solution, negative colloid is formed(B) Ultra centrifugation process is used for

preparing hydrophobic colloids(C) Milk can be coagulated by adding acidic

or basic solution to it.(D) Protein sols are not always positive sol



� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj

(D) gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi

ds vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa

ls fdlh ,d ds vuqlkj fn;s tk,axs %


dks dkyk fd;k gSA



1. dkSulh fof/k }kjk dksyksbM dk 'kqf¼dj.k fd;k tk ldrkgS -(A) la?kuu (B) isIVhdj.k(C) Ldanu (D) viksgu

2. dkSuls dFku lgh gS -(A) KI foy;u esa AgNO3 foy;u dk vkf/kD; foy;u

feykus ij ½.kkRed dksyksbM curk gS(B) tyfojks/kh dksyksbM cukus ds fy, vYVªk vidsUæ.k

izØe dk iz;ksx fd;k tkrk gS(C) nw/k esa vEyh; ;k {kkjh; foy;u feykdj Ldafnr

fd;k tk ldrk gSA(D) izksVhu lkWy ges'kk /kukRed lkWy ugha gksrs gSa

PART 2 - CHEMISTRY

TOPIC : Inorganic- Chemical bonding ; Organic- Aromatic Compounds, Biomolecules/Amino Acids , Polymer,Practical Organic Chemistry chemistry in everyday life ; Physical- Liquid Solution, Surface Chemistry

ALLEN


3. Identify incorrect statement -(A) A catalyst cannot effect the position of

equilibrium of a reaction(B) A catalyst may change pre-exponential

factor of a reaction(C) Heterogeneous catalytic reactions are

always zero ordered reactions(D) It is possible that a species may increases

activation energy of a reaction

4.

Cl

NO2

O N2 NO2

(1)KOH(2)H+¾¾¾¾® X

Correct option regarding X?(A) X is more acidic than H2CO3

(B) X has alcoholic OH group(C) IUPAC name of X is

1-hydroxy-2, 4,6-trinitrobenzene(D) Formation of X is example of

electrophilic aromatic substitution

5. RCH = O 3

3


+¾¾¾¾¾¾¾® RCH – NH3

CO2–

+

the amino acid which can not be preparedin this way is(A) Glycine (B) Alanine(C) Valine (D) Leucine

3. xyr dFku crkbZ;s -(A) mRizsjd vfHkfØ;k dh lkE; fLFkfr dks izHkkfor ugha

dj ldrk gS(B) mRizsjd vfHkfØ;k ds iwoZ pj&?kkrkadh xq.kkad dks

ifjofrZr dj ldrk gS(C) fo"kekaxh mRizsj.k vfHkfØ;k,sa ges'kk 'kw U; dksfV

vfHkfØ;k,sa gksrh gS(D) ;g lEHko gS fd ,d Lih'kht vfHkfØ;k dh lfØ;.k

ÅtkZ c<+k ldrh gS

4.

Cl

NO2

O N2 NO2

(1)KOH(2) H+¾¾¾¾® X

X ds lanHkZ es lgh fodYi gS\

(A) H2CO3 dh rqyuk es X vf/kd vEyh; gSA

(B) X es , sYdksgkWfyd OH lewg gSA

(C) X dk IUPAC uke 1-gkbMªkWDlh-2,4,6-VªkbZukbVªkscsathu

gSA

(D) X dk fuekZ.k bysDVªkWuLusgh ,sjksesfVd izfrLFkkiu dk

mnkgj.k gSA

5. RCH = O 3

3


+¾¾¾¾¾¾¾® RCH – NH3

CO2–

+

,slk vehuks vEy ftls bl fof/k ls cuk;k ugh tk ldrk

gS

(A) Xykbflu (B) ,sykfuu

(C) osfyu (D) Y;qflu

ALLEN


6.C

SNH

OO

O

For this compound if Lassaigne test iscarried out, what is the observation?(A) Blue colour is observed(B) Purple colour precipitate is observed(C) White precipitate is found which is

soluble in NH3 (excess)(D) Blood red colour is observed.

7. Which of the following have highest numberof 90° bond angle?

(A) SF6 (B) IF7 (C) PF5 (D) 4ICl-

8. Which of the following have intra molecularH-bonding and found as dimer form invapour phase.(A) H2O (B) Acetic acid(C) Fumaric acid (D) none of these

9. Which of the following oxy acid containmaximum number of ionizable H-atom?(A) Pyrosulphuric acid(B) Pyrophosphoric acid(C) Orthophosphoric acid(D) Pyrophosphorous acid

10. At 300K, vapour pressure of pure toluene &pure benzene are 40 and 120 torrrespectively. For a solution which boils at300K and 50 torr, % w/w of benzene insolution is-(A) 10.8% (B) 21.6% (C) 32.6% (D) 89.2%

6.C

SNH

OO

O

bl ;kSfxd ds fy;s ;fn ySlkXus ijh{k.k djk;k tkrkgS rks izs{k.k D;k gS\(A) uhyk jax izsf{kr gksrk gSA(B) cSxuh jax dk vo{ksi izsf{kr gksrk gSA(C) 'osr vo{ksi ik;k tkrk gS tks NH3 (vkf/kD;) esa

foys; gSA(D) jDr tSlk yky jax izsf{kr gksrk gSA

7. fuEu eas ls fdlesa 90° ds ca/k dks.kksa dh la[;k lokZfèkd

gS ?

(A) SF6 (B) IF7 (C) PF5 (D) 4ICl-

8. fuEu esa ls fdlesa vUr%vkf.od H-ca/ku mifLFkr gSa rFkk

ok"i voLFkk esa f}yd ds :i esa ik;k tkrk gS

(A) H2O (B) ,sflfVd vEy

(C) ¶;qesfjd vEy (D) bueas ls dksbZ ugha

9. fuEu es a ls dkSu ls vkWDlh vEy es a vk;uu ;k sX;

H-ijek.kqvksa dh vf/kdre la[;k mifLFkr gS?(A) ik;jkslY¶;qfjd vEy(B) ik;jksQkLQksfjd vEy(C) vkFkksZQkLQksfjd vEy(D) ik;jksQkLQksjl vEy

10. 300K ij 'kq¼ VkWyqbZu rFkk 'kq¼ csathu dk ok"i nkcØe'k% 40 rFkk 120 torr gSA ,d foy;u tks 300KrFkk 50 torr ij mcyrk gS] ds fy, foy;u esa casthudh ek=k % w/w gS&(A) 10.8% (B) 21.6% (C) 32.6% (D) 89.2%

ALLEN


11. For an aqueous solution of 0.1M Ba(NO3)2 ,osmotic pressure at 300K is found to be6.576 atm. Percentage dissociation ofBa(NO3)2 in solution is nearly-(A) 91.3% (B) 84%(C) 74% (D) 100%

12. Which of the following is not example ofartifical sweetner?(A) Alitame(B) Aspartame(C) Sucrolose(D) Chloroxylenol

13. Which of the following heterocyclic base isnot present in DNA?(A) Guanine (B) Adenine(C) Quinoline (D) Cytosine

14. Which of the following contain maximumionic character according to Hanny andSmith formula?(A) HF (B) HCl (C) HBr (D) HI

15. Which of the following does not form duringthe hydrolysis of XeF6?(A) XeOF4 (B) XeO3

(C) XeO2F2 (D) XeO4

16. A solution contains 31 gm of ethylene glycolin 200 gm of water. If this solution is cooledto –5.4ºC. what mass of ice will separate outat this temperature ?[Kf for water = 1.8 K-kgmol–1](A) 33.3 g (B) 37.1 g(C) 48.2 g (D) 62.1 g

11. 0.1M Ba(NO3)2 ds ,d tyh; foy;u ds fy, 300Kij ijklj.k nkc 6.576 atm ik;k x;kA foy;u esa

Ba(NO3)2 dk izfr'kr fo;kstu yxHkx gS

(A) 91.3% (B) 84%

(C) 74% (D) 100%

12. Ïf=e e/kwjd (sweetner) dk dkSulk mnkgj.k ugh gS\

(A) ,sfyVse

(B) ,sLikVsZe

(C) lqØksyksl

(D) DyksjksDlhysukWy

13. fuEu es ls dkSulk fo"kepØh; {kkj DNA es mifLFkrugh gS\(A) xqvkfuu (B) ,sMhfuu(C) Dohuksyhu (D) lk;Vksflu

14. gsuh rFkk fLeFk lw= ds vuqlkj] fuEu esa ls fdlesa] vfèkdre

vk;fud y{k.k mifLFkr gaS\

(A) HF (B) HCl (C) HBr (D) HI

15. XeF6 ds ty vi?kVu }kjk] fuEu eas ls dkSu lh Lih'khtizkIr ugha gksrh gS ?(A) XeOF4 (B) XeO3

(C) XeO2F2 (D) XeO4

16. ,d foy;u ftlesa 31 gm ,sfFkyhu XykbdkWy] 200 gmty esa mifLFkr gSA ;fn bl foy;u dks –5.4ºC rdBaMk fd;k tkrk gS rks bl rki ij cQZ dk fdruk æO;ekui`Fkd gksxk ?[ty ds fy, Kf = 1.8 K-kgmol–1](A) 33.3 g (B) 37.1 g(C) 48.2 g (D) 62.1 g

ALLEN


17.

NO2

NO2

3 22


++ - °

+

¾¾¾¾¾¾¾¾¾® Final Product

Final Product is?

(A)

Cl

(B)

Cl

Cl

(C)

NO2

Cl(D)

Cl

ClO N2

18. Which of the following species isparamagnetic and has bond order of 2.5

(A) 2NÅ (B) O2 (C) 22O - (D) 2CÅ

19.

NH2

22 4

32 4

–


D

¾¾¾¾¾¾¾® Product

Molecular formula of the major product is?(A) C6H5N2SO3 (B) C6H6N2O2

(C) C8H8N2O3 (D) C6H5NSO5

20. Which of the following allotrope of carbon

does not contain dangling bond?(A) Graphite (B) Diamond(C) Fullerene (D) None of these

17.

NO2

NO2

3 22


++ - °

+

¾¾¾¾¾¾¾¾¾® vfUre mRikn

vfUre mRikn gS\

(A)

Cl

(B)

Cl

Cl

(C)

NO2

Cl(D)

Cl

ClO N2

18. fuEu eas ls dkSu lh Lih'kht vuqpqEcdh; gS rFkk ftldk

ca/k Øe 2.5 gS?

(A) 2NÅ (B) O2 (C) 22O - (D) 2CÅ

19.

NH2

22 4

32 4

–


D

¾¾¾¾¾¾¾® mRikn

eq[; mRikn dk vkf.od lw= gS\(A) C6H5N2SO3 (B) C6H6N2O2

(C) C8H8N2O3 (D) C6H5NSO5

20. dkcZu ds] fuEu esa ls dkSu ls vij:i esa > wyk ca/k(dangling bond) mifLFkr ugha gS ?(A) xzsQkbV (B) ghjk(C) Qqyjhu (D) bueas ls dksbZ ugha

ALLEN





0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••


1. 0.5 M aq. solution of a weak base BOH isprepared. Find elevation of boiling point(DTb) of this solution ( in K) given.For BOH ; Kb = 2.5 × 10–2 [ionisation constant]For Water ; Kb = 5.2 K – Kg mol–1

[Ebullioscopic constant]Assume [Molality ; Molarity]

[kaM-II : (vf/kdre vad : 20)� bl [kaM esa ik¡p iz'u gSaA� iz R; sd iz'u dk mÙkj ,d l a[; k Red e ku

(NUMERICAL VALUE) gSA� iz R; sd i z'u ds mÙkj ds lgh la[; kRed eku

(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuks a rdVª ad sV@jkm aM vk WQ (truncate/round-off) djs a_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaAmnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk%&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. ,d nqcZy {kkj BOH dk 0.5 M tyh; foy;u cuk;kx;k gSA fn;s x;s bl foy;u dk DoFkukad esa mUu;u(DTb) (K esa) Kkr dhft,BOH ds fy, ; Kb = 2.5 × 10–2 [vk;uhdj.k fu;rkad]ty ds fy, ; Kb = 5.2 K – Kg mol–1

[bcqfyvksLdksfid fu;rkda]ekusa fd [eksyyrk ; eksyjrk]

ALLEN


2. Oxygen gas is maintained at partial pressureof 0.54 atm above water sample. Henry'sconstant for dissolution of O2 in water is 105

atm. If molality of O2 in water at equilibriumis (y × 10–x) final value of 'x'

3. Find the total number of allotropic formis/are molecular solid.

White phosphorous, Red phosphorous,Black phosphorous, Rhombic sulphur,Monoclinic sulphur, Diamond, Graphite,Fullerene.

4. How many aldose when reacts with excessAc2O / pyridine increases molecular weightby 168 unit?

(A) Xylose

(B) Galactose

(C) Glucose

(D) Erythrose

(E) Lyxose

(F) Ribose

(G) Arabinose

5. Find the total number of species havepermanant dipole moment.

CH3Cl, CH2Cl2, CHCl3, CCl4,PF2Cl3, PF3Cl2,PF4Cl, SO2, SO3

2. vkWDlhtu xSl dks ty ds uewus ds Åij 0.54 atm dsvkaf'kd nkc ij fu;af=r fd;k x;k gSA ty esa O2 dsfoyk;du ds fy, gsujh fu;rkad 105 atm gSA ;fn lkE;ij ty esa O2 dh eksyyrk (y × 10–x) gS rks 'x' dk ekuKkr dhft,

3. fuEu es ls ,sls vij:iks dh dqy la[;k crkbZ;s tks vkf.od

Bksl gS

'osr QkWLQksjl] yky QkWLQksjl] dkyk QkWLQksjl]

fo"keyEck{k lYQj] ,durk{k (Monoclinic) lYQj]

ghjk] xzsQkbV] Qqyjhu

4. fdrus ,sYMkst Ac2O / fijhMhu ds vkf/kD; ds lkFk fØ;k

djrs gS rks v.kqHkkj 168 bZdkbZ }kjk c<+ tkrk gS\

(A) tkbyksl

(B) xSysDVksl

(C) Xyqdksl

(D) bZjhFkzk sl

(E) ykblksl

(F) jkbcksl

(G) vjsfcuksl

5. fuEu es ls ,slh Lihf'kt dh dqy la[;k crkbZ;s tks LFkk;hf}/kz qo vk?kw.kZ j[krh gS\CH3Cl, CH2Cl2, CHCl3, CCl4, PF2Cl3, PF3Cl2,PF4Cl, SO2, SO3

ALLEN






1. Let <an>, <bn>, <cn> where n Î N arearithmetic progressions, such thata1 + b1 + c1 = 10 and a2 + b2 + c2 = 20 then thevalue ofa2019 + b2019 + c2019 equals to

(A) 20180 (B) 20190(C) 2018 (D) 2019

2. Equationx2 – 2(k – a)x + k2 + a2 – 16k – b + 12 = 0has repeated roots " k Î R, then value of(a + b) is :-(A) 8 (B) 12 (C) 16 (D) 20



� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj

(D) gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi

ds vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa

ls fdlh ,d ds vuqlkj fn;s tk,axs %


dks dkyk fd;k gSA



1. ekuk <an>, <bn>, <cn> tgk¡ n Î N lekUrj Js<+h;k¡bl çdkj gS fd a1 + b1 + c1 = 10 rFkk

a2 + b2 + c2 = 20 gks] rks (a2019 + b2019 + c2019)dk eku gksxk

(A) 20180 (B) 20190

(C) 2018 (D) 2019

2. lehdj.kx2 – 2(k – a)x + k2 + a2 – 16k – b + 12 = 0ds iqujko`Ùk ewy gks] rks (a + b) dk eku gksxk :-(A) 8 (B) 12(C) 16 (D) 20

PART 3 - MATHEMATICSTOPIC : Complex number, Matrices, Determinants, Quadratic Equations, Sequence & Series

ALLEN


3. The product of roots of l- + =2log2x 4x 4 0is 8 then l is :-(A) 2 (B) 2 2

(C) 3 (D) 3

4. If ( )n

2n r

r 1

1S t n 2n 9n 136=

= = + +å then value

of =

ån

rr 1

t is

(A) ( )+n n 1

2 (B) ( )+n n 2

2

(C) ( )+n n 3

2 (D) ( )+n n 5

2

5. A geometric progression has fourconsecutive positive terms a1, a2, a3, a4. If

=3

1

a 9a

and + =1 24a a3 , then a4 equals to

(A) 3 (B) 9 (C) 27 (D) 3 36. If 3p2 = 5p + 2 and 3q2 = 5q + 2, then the

equation whose roots are 3p – 2q and 3q – 2pis(A) x2 – 5x + 100 = 0(B) 3x2 – 5x – 100 = 0(C) 3x2 + 5x + 100 = 0(D) 5x2 – x + 7 = 0

7. If a, b are roots of equation x2 – 2x – 1 = 0,then value of 5a4 + 12b3 is(A) 153 (B) 169 (C) 183 (D) 168

3. l- + =2log2x 4x 4 0 ds ewyksa dk xq.kuQy 8 gks] rks ldk eku gksxk :-

(A) 2 (B) 2 2

(C) 3 (D) 3

4. ;fn ( )n

2n r

r 1

1S t n 2n 9n 136=

= = + +å gk s] rk s

=ån

rr 1

t dk eku gksxk

(A) ( )+n n 1

2 (B) ( )+n n 2

2

(C) ( )+n n 3

2 (D) ( )+n n 5

25. ,d x q.kk sÙ k j Js<+ h ds pkj Øekxr /kukRed in

a1, a2, a3, a4 gSA ;fn =3

1

a 9a

rFkk + =1 24a a3 gks] rks

a4 dk eku gksxk

(A) 3 (B) 9 (C) 27 (D) 3 36. ;fn 3p2 = 5p + 2 rFkk 3q2 = 5q + 2 gks] rks og

lehdj.k] ftlds ewy 3p – 2q rFkk 3q – 2p gks] gksxh

(A) x2 – 5x + 100 = 0(B) 3x2 – 5x – 100 = 0(C) 3x2 + 5x + 100 = 0(D) 5x2 – x + 7 = 0

7. ;fn lehdj.k x2 – 2x – 1 = 0 ds ewy a, b gks] rks5a4 + 12b3 dk eku gksxk(A) 153 (B) 169 (C) 183 (D) 168

ALLEN


8. If a, b, c are in H.P. then - -b b ba , , c2 2 2

are

in :-(A) A.P. (B) G.P.(C) A.G.P. (D) H.P.

9. If M and N are two invertible skewsymmetric matrices such that MN = NM then(NTM–1N–1)T is(A) M (B) M–1 (C) N (D) N–1

10. a and b are roots of equation x2 – 2x + 4 = 0where Im(a) < 0 then value of a15 + b10.z0 is

(where 2i3

0z e ,i 1p

= = - )

(A) 210 – 215 (B) 215 – 210

(C) 0 (D) 111. If < li , mi, ni > are direction cosines of three

mutually perpendicular straight lines for

i = 1,2,3 and 1 1 1

2 2 2

3 3 3

m nL m n

m n


l

l

l

, then absolute

value of det(2L–1) is

(A) 8 (B) 18 (C) 4 (D)

14

12. Let A and B are square matrices of order3 × 3 such that AB = I, adjA = B – A, then(A) |B| > 1(B) 0 < |B| < 1(C) |A| > 1(D) |A| < 0(where |X| denotes determinant value ofmatrix X)

8. ;fn a, b, c gjkRed Js<+h esa gks] rks - -b b ba , , c2 2 2

fuEu Js<+h esa gksaxs :-(A) lekUrj Js<+h (B) xq.kksÙkj Js<+h(C) lekUrj xq.kksÙkj Js<+h (D) gjkRed Js<+h

9. ;fn M rFkk N nks O;qRØe.kh; fo"ke lefer vkO;wg blçdkj gS fd MN = NM gks] rks (NTM–1N–1)T gksxk(A) M (B) M–1

(C) N (D) N–1

10. ;fn a rFkk b lehdj.k x2 – 2x + 4 = 0 ds ewy gSa tgk¡Im(a) < 0 gks] rks a15 + b10.z0 dk eku gksxk

(tgk¡ 2i3

0z e ,i 1p

= = - )

(A) 210 – 215 (B) 215 – 210

(C) 0 (D) 111. ;fn i = 1, 2, 3 ds fy;s rhu ijLij yEco~r ljy js[kkvksa

dh fnd~dk sT;k,¡ < l i , m i, ni > g Sa rFk k

1 1 1

2 2 2

3 3 3

m nL m n

m n


l

l

l

gks] rks det(2L–1) dk fujis{k eku

gksxk

(A) 8 (B) 18 (C) 4 (D)

14

12. ekuk A rFkk B; 3 × 3 dksVh ds oxZ vkO;wg bl çdkj gSfd AB = I, adjA = B – A gks] rks(A) |B| > 1(B) 0 < |B| < 1(C) |A| > 1(D) |A| < 0(tgk¡ |X|, vkO;wg X ds lkjf.kd eku dks n'kkZrk gS)

ALLEN


13. Let A be a square matrix of order 2 × 2 suchthat tr(A) = –2, A2 = I, then |A–1 + A2| is(where |X| and tr(X) denote determinantvalue and trace of matrix X respectively)(A) 1(B) –1(C) 0(D) can't be determined

14. 2 21 1 1

1 1+ +

- w w - w w - is equal to (w is non

real cube root of unity)(A) 0 (B) 1 (C) –1 (D) w

15. If ( ) ( )n

2r 1

1 10r r 1 1 10=

w=

w - w - + - wå

then n is 2i3where e ,i 1pæ ö

w = = -ç ÷è ø

(A) 9 (B) 10 (C) 8 (D) 11

16. The system of equationsax + y + z = ax + ay + z = ax + y + az = ahas no solution, if(A) a = –2, only (B) a = –2, – 1 only(C) a = –2,1 only (D) a Î R – {–2}

17. Let A = [aij]3×3 and C = [cij]3×3 " i, j Î {1,2,3}be two square matrices, where cij is cofactorof aij in |A|such that 3cij – 2aij = 0 andATA = lI, l ¹ 0, then value of l is

(A) 23 (B)

49 (C)

94 (D)

32

13. ekuk A, 2 × 2 dksVh dk oxZ vkO;wg bl çdkj gS fdtr(A) = –2, A2 = I gks] rks |A–1 + A2| gksxk(tgk¡ |X| rFkk tr(X) Øe'k% vkO;wg X ds lkjf.kd ekurFkk vuqjs[k dks n'kkZrk gS)(A) 1(B) –1(C) 0(D) Kkr ugha fd;k tk ldrk gSA

14. 2 21 1 1

1 1+ +

- w w - w w - gk sxk (w, bdkbZ dk

vokLrfod ?ku ewy gS)(A) 0 (B) 1 (C) –1 (D) w

15. ;fn ( ) ( )n

2r 1

1 10r r 1 1 10=

w=

w - w - + - wå

gks] rks n gksxk 2i3e ,i 1pæ ö

w = = -ç ÷è ø

tgk¡

(A) 9 (B) 10 (C) 8 (D) 1116. lehdj.k fudk;

ax + y + z = ax + ay + z = ax + y + az = adk dksbZ gy ugha gksxk] ;fn

(A) dsoy a = –2 gks (B) dsoy a = –2, – 1 gks(C) dsoy a = –2,1 gks (D) a Î R – {–2} gks

17. ekuk A = [aij]3×3 rFkk C = [cij]3×3 " i, j Î {1,2,3} nksoxZvkO;wg gS] tgk¡ |A| esa cij , aij dk lg[k.M bl izdkjgS fd 3cij – 2aij = 0 rFkk ATA = lI, l ¹ 0 gks] rks l dkeku gksxk

(A) 23 (B)

49 (C)

94 (D)

32

ALLEN


18. If

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( )

2 22 2 3

2 2 2 6 22 3 4 5

2 2 23 4 5

1 1 1

1 1 1 1 1

1 1 1

a + a + a +

a + a + a + = la - a + a

a + a + a +

,

then l is equal to

(A) 1 (B) 4

(C) 2 (D) –2

19. The following system of linear equations

x + y + z = 1

x + ay + z = 1

x + by + az = 0

has infinite solutions, then

(A) b has only one value

(B) b Î f

(C) b Î R

(D) b Î R – {1}

20. If a,b,g are the roots of the equation

x3+ax2 + b = 0, b ¹ 0 and

21 1 1

1 1 1

1 1 1

a b g

D =b g a

g a b

, then

D is equal to(A) a2 – b (B) b2 – a

(C) a2b2 (D) 0

18. ;fn

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( )

2 22 2 3

2 2 2 6 22 3 4 5

2 2 23 4 5

1 1 1

1 1 1 1 1

1 1 1

a + a + a +

a + a + a + = la - a + a

a + a + a +

gks] rks l gksxk

(A) 1 (B) 4

(C) 2 (D) –2

19. fuEu js[kh; lehdj.k fudk;

x + y + z = 1

x + ay + z = 1

x + by + az = 0

ds vuar gy gks] rks

(A) b dk dsoy ,d eku gksxk

(B) b Î f

(C) b Î R

(D) b Î R – {1}

20. ;fn a,b,g lehdj.k x3+ax2 + b = 0, b ¹ 0 ds ewy gSa

rFkk

21 1 1

1 1 1

1 1 1

a b g

D =b g a

g a b

gks] rks D dk eku gksxk

(A) a2 – b (B) b2 – a

(C) a2b2 (D) 0

ALLEN





0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••


1. Let 1 0 0

A 2 1 02 2 1


, then det(A10 + A–10) is

equal to

[kaM-II : (vf/kdre vad : 20)� bl [kaM esa ik¡p iz'u gSaA� iz R; sd iz'u dk mÙkj ,d l a[; k Red e ku

(NUMERICAL VALUE) gSA� iz R; sd i z'u ds mÙkj ds lgh la[; kRed eku

(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkku gS]rks la[;kRed eku dks n'keyo ds nk s LFkkuk s a rdVª ad sV@jkm aM vk WQ (truncate/round-off) djs a_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaAmnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk%&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. ekuk 1 0 0

A 2 1 02 2 1


gks] rks det(A10 + A–10) dk eku

gksxk

ALLEN


2. Radius of 1Re 1i z

æ ö =ç ÷+è ø is r, then 3r

2 is equal to

(where z = x + iy, i 1= - , x,y Î R)

3. Let z1 and z2 are two complex numbers such

that 1 2z z 1= and arg(z1) – arg(z2) = 23p

, then

value of4i3

1 2z z ep

+ is equal to

(where i 1= - , –p < arg(z) £ p)

4. All the points in the set

t 2iS ,t R,i 1t 2i

+ì ü= Î = -í ý-î þlie on the curve

ƒ(z) = 0, z = x + iy where x, y Î R

then maximum value of |z + i – 2| is5. If z1 and z2 are non zero solutions of equation

2z z iz+ = where i 1= - , then

|z1 + z2| is equal to

2. 1Re 1i z

æ ö =ç ÷+è ø dh f=T;k r gks] rks 3r

2 gksxk

(tgk¡ z = x + iy, i 1= - , x,y Î R)

3. ekuk z1 rFkk z2 nks lfeJ la[;k;sa bl çdkj gS fd

1 2z z 1= rFkk arg(z1) – arg(z2) = 23p

gks] rks

p

+4i3

1 2z z e dk eku gksxk

(tgk¡ i 1= - , –p < arg(z) £ p)

4. leqPp;

t 2iS ,t R,i 1t 2i

+ì ü= Î = -í ý-î þ es a lHkh fcUnq oØ

ƒ(z) = 0, z = x + iy ij fLFkr gS tgk¡ x, y Î R gks] rks

|z + i – 2| dk vf/kdre eku gksxk

5. ;fn lehdj.k 2z z iz+ = (tgk¡ i 1= - ) ds v'kwU;

gy z1 rFkk z2 gks] rks |z1 + z2| dk eku gksxk

ALLEN


SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg

Page 28/28 05112019Enthusiast Course/Score-I/Paper-2 1001CJA102119029

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